$$\eqalign{ & {\text{Solving, }}x = \frac{{3{c^2} - 8}}{{9{c^2} - 4c - 4}},\,y = \frac{{5 - c}}{{9{c^2} - 4c - 4}} \cr & {\text{Now,}}\,\,\mathop {\lim }\limits_{c \to 1} \frac{{3{c^2} - 8}}{{9{c^2} - 4c - 4}} = \frac{{ - 5}}{1} = - 5\,\,{\text{and}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {\lim }\limits_{c \to 1} \frac{{5 - c}}{{9{c^2} - 4c - 4}} = \frac{4}{1} = 4 \cr} $$

The cartesian equation is $$x + 2 = \frac{{y - 1}}{3},\,{\text{i}}{\text{.e}}{\text{.}},\,3x - y + 7 = 0.$$

Clearly, $$OP = OQ = 1$$    and $$\angle QOP = \alpha - \theta - \theta = \alpha - 2\theta $$

The bisector of $$\angle QOP$$   will be perpendicular to $$PQ$$  and also bisect it. Hence, $$Q$$ is the reflection of $$P$$ in the line $$OM$$  which makes an angle equal to $$\angle MOP + \angle POX$$     with the $$x$$-axis, i.e., $$\frac{1}{2}\left( {\alpha - 2\theta } \right) + \theta = \frac{\alpha }{2}$$
So that slope of $$OM$$ is $$\tan \left( {\frac{\alpha }{2}} \right).$$

If the line cuts off the axes at $$A$$ and $$B,$$ then the area of triangle is
$$\eqalign{ & \frac{1}{2} \times OA \times OB = T \cr & {\text{or }}\frac{1}{2} \times a \times OB = T \cr & {\text{or }}OB = \frac{{2T}}{a} \cr} $$
Hence, the equation of line is $$\frac{x}{{ - a}} + \frac{y}{{\frac{{2T}}{a}}} = 1\,\,{\text{or }}2Tx - {a^2}y + 2aT = 0$$

$$\eqalign{ & y - \sqrt 3 x - 5 = 0{\text{ line one}} \cr & \sqrt 3 y - x + 6 = 0{\text{ line two}} \cr & y = mx + c \cr & y = \sqrt 3 x + 5 \cr & y = \frac{x}{{\sqrt 3 }} - \frac{6}{{\sqrt 3 }} \cr & {m_1} = \sqrt 3 \cr & {m_2} = \frac{1}{{\sqrt 3 }} \cr} $$
Angle between two lines,
$$\eqalign{ & \tan \,\theta = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| = \left| {\frac{{\sqrt 3 - \frac{1}{{\sqrt 3 }}}}{{1 + \sqrt 3 \frac{1}{{\sqrt 3 }}}}} \right| = \frac{1}{{\sqrt 3 }} = \tan \,{30^ \circ } \cr & \therefore \,\theta = {30^ \circ } \cr} $$

The equation of the graph is $$y = \cos \,x.\cos \left( {x + 2} \right) - {\cos ^2}\left( {x + 1} \right)$$
$$\eqalign{ & {\text{or, }}y = \frac{1}{2}\left\{ {\cos \,2 + \cos \,2\left( {x + 1} \right)} \right\} - \frac{1}{2}\left\{ {1 + \cos \,2\left( {x + 1} \right)} \right\} \cr & = \frac{1}{2}\left( {\cos \,2 - 1} \right) \cr & = - \frac{1}{2}.2{\sin ^2}1 \cr & = - {\sin ^2}1 \cr} $$
The graph is parallel to the $$x$$-axis, and $$\left( {\frac{\pi }{2},{\mkern 1mu} - {{\sin }^2}1} \right)$$   satisfies it.

Given : The coordinates of points $$P, \,Q, \,R$$   are $$\left( { - 1,\,0} \right),\,\left( {0,\,0} \right),\,\left( {3,\,3\sqrt 3 } \right)$$     respectively.

$$\eqalign{ & {\text{Slope of }}QR = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{3\sqrt 3 }}{3} \cr & \Rightarrow \tan \,\theta = \sqrt 3 \cr & \Rightarrow \theta = \frac{\pi }{3} \cr & \Rightarrow \angle RQX = \frac{\pi }{3} \cr & \therefore \angle RQP = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3} \cr} $$
Let $$QM$$  bisects the $$\angle PQR,$$
$$\therefore $$ Slope of the line $$QM = \tan \frac{{2\pi }}{3} = - \sqrt 3 $$
$$\therefore $$ Equation of line $$QM$$  is $$\left( {y - 0} \right) = - \sqrt 3 \left( {x - 0} \right)$$
$$\eqalign{ & \Rightarrow y = - \sqrt 3 x\,\, \cr & \Rightarrow \sqrt 3 \,x + y = 0 \cr} $$